[Math] Bounding the matrix norm of a commutator $[A,B]$ in terms of the norms of $A$ and $B$

inequalitieslinear algebramatricesnormsoperator-norms

The setup is as in this question:
Given a norm $N$ over ${\bf M}_n(\mathbb C)$, it is a natural question to find the best constant $C_N$ such that
$$N([A,B])\le C_N N(A)N(B),\qquad\forall A,B\in{\bf M}_n(\mathbb C).$$

Equivalently, $C_N$ is the maximum of $N(AB-BA)$ provided that $N(A)=N(B)=1$.

Given examples of $C_N$ are

$C_N=\sqrt{2}$ if $N$ is the Frobenius norm
$C_N=2$ if $N$ is the operator norm $\| \cdot\|_2$
$C_N=4$ if $N$ is the numerical radius $r(A)=\sup\limits_{x\ne0}\dfrac{|x^*Ax|}{\|x\|^2}$ (See this answer to an MO question).

if $N$ is the induced $p$-norm, defined for $1\le p\le\infty$ by $\|A\| _p = \sup \limits _{x \ne 0} \frac{\| A x\| _p}{\|x\|_p}$, we have $C_N=2$ for $p=\infty$ (with $\|A\|_\infty $ being just the maximum absolute row sum of the matrix). Indeed, the lower bound $2$ for $\|\cdot\|_\infty $ is obtained by taking e.g. $A=\begin{pmatrix} 1&0\\1&0\end{pmatrix}$ and $B=\begin{pmatrix} 0&1\\0&-1\end{pmatrix}$, and it should be easy to prove that $2$ is also the general upper bound for $\|\cdot\|_\infty $.
Similarly, $C_N=2$ for $p=1$ (with $\|A\|_1 $ being the maximum absolute column sum of the matrix).

Knowing that $C_N\equiv2$ for $p=1,2,\infty$, is it true that the same holds for the induced $p$-norms for all $p\ge1$?

If $N$ runs over all possible matrix norms, what is the range of $C_N$? In particular, is it bounded below and/or above?

(To avoid trivialities, let's keep it homogeneous by only considering "normalized" norms, i.e. require $N(I_n)=1$. This does not seem to be part of the standard definition of a norm.)

Best Answer

A somewhat more general setting, namely, finding the best constant $C_{p,q,r}$ in \begin{equation*} \|AB-BA\|_p \le C_{p,q,r}\|A\|_q\|B\|_r, \end{equation*} for Schatten $p$,$q$,$r$-norms, is studied in this paper.

EDIT (1st Mar'17). See also this recent paper (LAA, 521(15), May 2017, Pages 263–282) that provides sharp bounds on the constant $C_{p,q,r}$ above for real matrices.

Related Solutions

[Math] Norm of Cesaro matrix

Here there is likely no closed form for $\|C_n\|_2$, but I can show $\|C_n\|_2 > 2 - O(1/\log n)$; given that also $\|C_n\|_2 \leq 2$ (is this easy to prove?) it follows that $\|C_n\|_2 \rightarrow 2$ as $n \rightarrow \infty$.

In general we can consider, for fixed $p \in [1,\infty]$, the $l^p$-operator norm of the Cesàro transformation $C_n$ that takes any finite sequence $(a_1,a_2,\ldots,a_n)$ to its sequence of averages $j^{-1} \sum_{i=1}^j a_i$. [If I've kept my $i$'s and $j$'s straight, that's the action of $C_n$ on row vectors, so this operator norm is what you'd call $\|C_n\|_q$ where $p^{-1} + q^{-1} = 1$; fortunately for $p=2$ the $l^p$ norm is self dual.] I claim that this operator norm is at least $p - O(1/\log n)$. Indeed Let $v_n(p)$ be the sequence whose $j$-th term is $j^{-1/p}$, so $\|v_n\|_p^p = H_n$. Then the image of $v_n(p)$ under $C_n$ differs from $p v_n(p)$ by a vector of $l_p$-norm $O(1)$, because its $j$-th entry is between $$j^{-1} \int_1^j \phantom. x^{-1/p} \phantom. dx = p \cdot (j^{-1/p} - j^{-1})$$ and $$j^{-1} \int_0^j \phantom. x^{-1/p} \phantom. dx = p \phantom. j^{-1/p}.$$ We have thus found a nonzero vector whose norm is multiplied by $p - O(1/\log n)$, which proves the claimed lower bound.

I don't expect that $\|C_n\|_2$ has a closed form, because it is a root of the characteristic polynomial of $C_n^{\rm T} C_n^{\phantom.}$, and these characteristic polynomials have maximal Galois group for small $n$: running the gp code

C(n) = matrix(n,n,i,j,(i<=j)/j)
S2(M) = charpoly(M*M~)
P(n)=S2(C(n))
for(n=1,11,print(polgalois(P(n))))

I found that the Galois group is $S_n$ for each $n \leq 11$. The norm $\|C_n\|_2$ can then also be computed (as long as $n$ is small enough that the computation terminates in reasonable time) by telling gp

N(n) = sqrt(vecmax(real(polroots(P(n)))))

and the convergence to $2$ seems slow enough that $O(1/\log n)$ might well be the correct error estimate.

Curiously it seems that the action of $C_n$ on the column vector $(a_n(2))^{\rm T}$ comes rather close to the actual $l^2$-operator norm. For example, when $n=40$ this yields the lower bound $1.5520\ldots$ on the norm (which is actually $1.5594\ldots$), while the row vector gets only $1.4201\ldots$. If this persists as $n \rightarrow \infty$ it should yield a much better estimate on $\|C_n\|_2$.

[Math] Hölder’s inequality for matrices

There are (at least two) "generalizations" of Hölder inequality to the non-commutative case. One is the so called tracial matrix Hölder inequality:

$$ |\langle A, B \rangle_{HS} |= |\mathrm{Tr} (A^\dagger B) | \le \| A\|_p \,\, \| B\|_q $$

where $\| A\|_p$ is the Schatten $p$-norm and $1/p+1/q=1$. You can find a proof in Bernhard Baumgartner, An Inequality for the trace of matrix products, using absolute values.

Another generalization is very similar to what you wrote and reads

$$ \parallel|AB|\parallel \, \le\, \parallel |A|^p\parallel^{1/p} \,\, \parallel|B|^q \parallel^{1/q} $$ where $|M|:=(M^\dagger M)^\frac12$ and it holds whenever $ \parallel \cdot \parallel$ is a unitarily invariant norm. You can find a proof in the book of Bhatia Matrix Analysis.

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